I'm confused how to prove that $\lim_{x\rightarrow 1} \sin (x^2 - 1) / (x^2 - 1) = \lim_{x\rightarrow 0} \sin(x)/x$.
Assuming $\lim_{x\rightarrow 1} \sin (x^2 - 1) / (x^2 - 1) = l$, then there's $\delta > 0$ s.t. for all $x$, if $0 < |x - 1| < \delta$ implies that $ | \sin (x^2 - 1) / (x^2 - 1) - l| < \epsilon $ for all $\epsilon > 0$.
My approach: if we take $|x| < \delta$, then $| (x + 1) - 1| < \delta$ this implies that $| \sin (x^2 - 2x) / (x^2 - 2x) - l| < \epsilon$, which probably isn't the route I need to take.
Intuitively it makes sense, since $x^2 - 1$ approaches $0$, then we should be able to replace it with its limit.
This is for the theorem of composition since $f(x)=x^2-1 \to 0$ and $x^2-1 \neq 0$ for $x\neq 1$ the following holds
$$\lim_{x\rightarrow 1} \frac{\sin (f(x)) }{ f(x)} = \lim_{x\rightarrow 0} \frac{\sin(x)}x=1$$
Refer also to the related